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Elanso [62]
3 years ago
13

To test the effects of a new fertilizer, 100 plots were divided in half. Fertilizer A is randomly applied to one half, and B to

the other. This is ;
(a) an observational study.
(b) a matched pairs experiment.
(c) a completely randomized experiment.
(d) a block design, but not a matched pairs experiment.
(e) impossible to classify unless more details of the study are provided
Engineering
2 answers:
Kruka [31]3 years ago
7 0

Answer:

(b) a matched pairs experiment.

Explanation:

Matched pairs experiment are a special case of a random block design and is usually used when the experiment only has two conditions. In this case the conditions are the different fertilizers.

Oduvanchick [21]3 years ago
3 0

Answer: (b) a matched pairs experiment.

Explanation:

A match pair design or experiment is a randomized experimental procedure. It can be used in the experiment with two different sets of conditions and the subjects can be grouped into pairs. It is based on some of the blocking variable. The subjects are randomly assigned to different treatments.  

The given situation is an example of a matched pair experiment. This is because of the fact that there are two sets of conditions that is the use of fertilizer A and fertilizer B on different halves of the plot randomly to determine their effects.

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Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which
Rudik [331]

ANSWERS:

-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor x_{1} =1

The temperature T_{1} =0^0 F

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a)

-V_{2} =2V_{1} ( double)

b)

-V_{2} =.5V_{2} (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb

then the state exists in the supper heated region.

a) from standard data

-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF

at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg

at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg

assume linear interpolation

\frac{P_{x}-P_{2(b)}  }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)}  }{v_{x}-v_{y}  }

P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y}  }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2

b)

-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}

from standard data

-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}

then the state exist in the wet zone

-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )

x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%

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